Chapter 7, Problem 69P

To determine

The speed and direction of the tangled mess of metal.

Answer to Problem 69P

The speed and direction of the car is $6.0\text{m/s}$ at $21°$ S of E.

Explanation of Solution

The figure 1 shows the final position of cars moved together into the fourth quadrant.

The collision betweenthe two cars is perfectly inelastic, so the final velocities of the cars are identical.

Write the equation for initial momentum of the object along x axis.

$p_{ix}=m_1{v}_{1ix}+m_2{v}_{2ix}$ (I)

Here, $p_{ix}$ is the initial momentum of the car before collsion along x direction, $m_1$ is the mass of the car, $v_{1ix}$ is the initial velocity of the car along north and east direction,  is the mass of the smaller car, and $v_{2ix}$ is the initial velocity of the smaller car along north and east direction.

Write the equation for final momentum of the object along x axis.

$p_{fx}=\left(m_1+m_2\right)v_{fx}$ (II)

Here, $p_{fx}$ is the final momentum of the car during collsion along x direction and $v_{fx}$ is the final velocity of the car.

Write the equation for initial momentum of the object aong y axis.

$p_{iy}=m_1{v}_{1iy}+m_2{v}_{2iy}$ (III)

Here, $p_{iy}$ is the initial momentum of the car after collsion along y direction, $v_{1iy}$ is the initial velocity of the car along north direction, and $v_{2iy}$ is the initial velocity of the smaller car along north direction.

Write the equation for final momentum of the object along y axis.

$p_{fy}=\left(m_1+m_2\right)v_{fy}$ (IV)

Here, $p_{fy}$ is the final momentum of the car along y direction and $v_{fy}$ is the final velocity of the car.

Rewrite the equation (I) by substituting 0 for $v_{2ix}$.

$p_{ix}=m_1{v}_{1ix}$ (V)

Rearrange the equation (II) and (V) for conservation of momentum to solve for $v_{fx}$.

$\begin{array}{c}{p}_{ix}=p_{fx}\\ m_1{v}_{1ix}=\left(m_1+m_2\right)v_{fx}\\ v_{fx}=\frac{m_1}{\left(m_1+m_2\right)}{v}_{1i}\cos \theta \end{array}$ (VI)

Rearrange the equation (III) and (IV) for conservation of momentum to solve for $v_{fy}$.

$\begin{array}{c}{p}_{iy}=p_{fy}\\ m_1{v}_{1iy}+m_2{v}_{2iy}=\left(m_1+m_2\right)v_{fy}\\ v_{fy}=\frac{m_1{v}_{1i}\sin \theta +m_2{v}_{2iy}}{\left(m_1+m_2\right)}\end{array}$ (VII)

Write the equation for final speed of the cars.

$v_f=\sqrt{v_{fx}^2+v_{fy}^2}$ (VIII)

Write the equation for direction of the cars.

$\theta ={\tan }^{-1}\left(\frac{v_{fy}}{v_{fx}}\right)$                                                                         (IX)

Conclusion:

Substitute $1700\text{kg}$ for $m_1$, $1300\text{kg}$ for $m_2$, $45°$ for $\theta$, and $14\text{m/s}$ for $v_{1i}$ in equation (VI).

$\begin{array}{c}{v}_{fx}=\frac{1700\text{kg}}{\left(1700\text{kg}+1300\text{kg}\right)}\left(14\text{m/s}\right)\cos 45°\\ =\frac{1700\text{kg}}{3000\text{kg}}\left(9.9\text{m/s}\right)\\ =5.61\text{m/s}\end{array}$

Substitute $1700\text{kg}$ for $m_1$, $1300\text{kg}$ for $m_2$, $45°$ for $\theta$, $-18\text{m/s}$ for $v_{2iy}$, and $14\text{m/s}$ for $v_{1i}$ in equation (VII).

$\begin{array}{c}{v}_{fy}=\frac{\left(1700\text{kg}\right)\left(14\text{m/s}\right)\sin 45°+\left(1300\text{kg}\right)\left(-18\text{m/s}\right)}{\left(1700\text{kg}+1300\text{kg}\right)}\\ =\frac{16829\text{kg}\cdot \text{m/s}-23400\text{kg}\cdot \text{m/s}}{3000\text{kg}}\\ =-2.19\text{m/s}\end{array}$

Substitute $-2.19\text{m/s}$ for $v_{fy}$ and $5.61\text{m/s}$ for $v_{fx}$ in equation (VIII).

$\begin{array}{c}{v}_f=\sqrt{{\left(5.61\text{m/s}\right)}^2+{\left(-2.19\text{m/s}\right)}^2}\\ =6.0\text{m/s}\end{array}$

Substitute $-2.19\text{m/s}$ for $v_{fy}$ and $5.61\text{m/s}$ for $v_{fx}$ in equation (IX).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-2.19\text{m/s}}{5.61\text{m/s}}\right)\\ ={\tan }^{-1}\left(-0.390\right)\\ =-21.3°\\ =-21°\end{array}$

Therefore, the speed and direction of the car is $6.0\text{m/s}$ at $21°$ S of E.